Vector Proof for Collinearity Using Given Segment Relations

Key takeaway: These problems look long, but the algebra is tidy once you express every vector from the same starting point. The collinearity proof usually falls out at the end as a scalar multiple.

Use the given information:

$\overrightarrow{PQ}=a,\quad \overrightarrow{SR}=2a,\quad \overrightarrow{SP}=b,\quad ST=2TQ.$

a) Find the requested vectors

i) $\overrightarrow{SQ}$

Since

$\overrightarrow{SQ}=\overrightarrow{SP}+\overrightarrow{PQ},$

we get

$\overrightarrow{SQ}=b+a.$

ii) $\overrightarrow{TQ}$

Given $ST=2TQ$, point $T$ divides $SQ$ in the ratio $2:1$. So

\qquad \overrightarrow{TQ}=\frac{1}{3}\overrightarrow{SQ}.$$ Hence $$\overrightarrow{TQ}=\frac13(a+b).$$ #### iii) $\overrightarrow{RQ}$ Go from $R$ to $S$ and then to $Q$: $$\overrightarrow{RQ}=\overrightarrow{RS}+\overrightarrow{SQ}.$$ Now $\overrightarrow{RS}=-\overrightarrow{SR}=-2a$, so $$\overrightarrow{RQ}=-2a+(a+b)=b-a.$$ #### iv) $\overrightarrow{PT}$ $$\overrightarrow{PT}=\overrightarrow{PS}+\overrightarrow{ST}$$ with $\overrightarrow{PS}=-b$ and $\overrightarrow{ST}=\frac23(a+b)$, so $$\overrightarrow{PT}=-b+\frac23(a+b)=\frac23a-\frac13b.$$ #### v) $\overrightarrow{TR}$ $$\overrightarrow{TR}=\overrightarrow{TS}+\overrightarrow{SR}$$ where $\overrightarrow{TS}=-\overrightarrow{ST}=-\frac23(a+b)$. Thus $$\overrightarrow{TR}=-\frac23(a+b)+2a=\frac43a-\frac23b.$$ ### b) Show that $P$, $T$, and $R$ are collinear Compare $\overrightarrow{PT}$ and $\overrightarrow{TR}$: $$\overrightarrow{TR}=\frac43a-\frac23b=2\left(\frac23a-\frac13b\right)=2\overrightarrow{PT}.$$ So $\overrightarrow{TR}$ is a scalar multiple of $\overrightarrow{PT}$. Therefore, $P$, $T$, and $R$ are collinear. $$\boxed{\overrightarrow{SQ}=a+b,\; \overrightarrow{TQ}=\frac13(a+b),\; \overrightarrow{RQ}=b-a,\; \overrightarrow{PT}=\frac23a-\frac13b,\; \overrightarrow{TR}=\frac43a-\frac23b}$$ --- **Pitfalls the pros know** 👇 Two things usually go wrong here. First, students flip $\overrightarrow{RS}$ and $\overrightarrow{SR}$. Second, the ratio $ST=2TQ$ means the whole segment $SQ$ is split into 3 equal parts, not 2. That is why $TQ=\frac13 SQ$. **What if the problem changes?** If the ratio were changed to $ST=k\,TQ$, then $T$ would divide $SQ$ in the ratio $k:1$, so $\overrightarrow{TQ}=\frac{1}{k+1}\overrightarrow{SQ}$ and $\overrightarrow{ST}=\frac{k}{k+1}\overrightarrow{SQ}$. The rest of the proof would follow the same pattern. `Tags`: collinearity proof, vector addition, section ratio